Since school we learn that “the order of the terms does not change the sum”. For finite sums, it is true. For infinite sums, it is not, and the clearest example appears in the same habilitation work of 1854 in which Riemann defined the integral.

A series that adds up to ln 2

The alternating harmonic series

$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots$$

converges, and its sum is $\ln 2 \approx 0.6931$. But there is something fragile about it. If we separate the positive terms from the negative ones, each group on its own is infinite:

$$1 + \frac{1}{3} + \frac{1}{5} + \cdots = \infty, \qquad \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots = \infty.$$

The series only converges because the positive and negative terms alternate and keep cancelling. A series like this, which converges but whose sum of absolute values is infinite, is called conditionally convergent.

Any number

Riemann realized that, in a conditionally convergent series, the terms can be reordered to obtain any sum. Choose a target, say 1.5, and follow the recipe:

  1. add positive terms, $1 + \frac{1}{3} + \frac{1}{5} + \cdots$, until you pass 1.5;
  2. add negative terms, $-\frac{1}{2} - \frac{1}{4} - \cdots$, until you are below 1.5;
  3. go back to the positive terms not yet used, and repeat.

Since each group on its own is infinite, there are always enough terms to pass the target again. And since the terms keep getting smaller, the oscillations around the target shrink. The sum of the reordered series is exactly 1.5. Each term of the original series appears exactly once; only the order has changed.

By the same recipe, the series can be made to diverge to $+\infty$ or to $-\infty$.

Exercise: one positive, two negatives

Reorder the alternating harmonic series by taking one positive term followed by two negative ones:

$$1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \frac{1}{12} + \cdots$$

Show that the sum is $\frac{1}{2}\ln 2$.

Group each positive term with the first negative one that follows it: $\left(1 - \frac{1}{2}\right) - \frac{1}{4} + \left(\frac{1}{3} - \frac{1}{6}\right) - \frac{1}{8} + \left(\frac{1}{5} - \frac{1}{10}\right) - \cdots$. Each bracket is half the positive term: $\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} - \cdots = \frac{1}{2}\left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots\right) = \frac{1}{2}\ln 2$. The same terms, in another order, give half the sum.

Exercise: the first steps towards the target 1.5

Applying the recipe with target 1.5, how many positive terms are added before the first negative one?

$1 = 1$; $1 + \frac{1}{3} \approx 1.333$; $+ \frac{1}{5} = 1.533$, which already passes 1.5. That is three positive terms ($1, \frac{1}{3}, \frac{1}{5}$). Then comes $-\frac{1}{2}$, which brings the sum to $1.033$, below 1.5, and the recipe goes back to the positive terms: $\frac{1}{7}, \frac{1}{9}, \ldots$

Sources: Bernhard Riemann, “Über die Darstellbarkeit einer Function durch eine trigonometrische Reihe” (1854, published 1867), section 3 · Peter Gustav Lejeune Dirichlet, “Beweis des Satzes, dass jede unbegrenzte arithmetische Progression…” (1837) · Konrad Knopp, Theory and Application of Infinite Series (1928).

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